Math Problems with Solutions for Grade 10

a) Maximum height occurs at the vertex of the parabola: \[ t = \frac{-b}{2a} = \frac{-15}{2(-5)} = \frac{15}{10} = 1.5\ \text{seconds} \] b) Substitute \( t = 1.5 \) into the equation to obtain the maximum height: \[ h(1.5) = -5(1.5)^2 + 15(1.5) + 20 = -11.25 + 22.5 + 20 = 31.25\ \text{meters} \] c) Set \( h(t) = 0 \): \[ -5t^2 + 15t + 20 = 0 \Rightarrow t^2 - 3t - 4 = 0 \] Factor: \[ (t - 4)(t + 1) = 0 \Rightarrow t = 4\ \text{or}\ t = -1 \] Time cannot be negative, hence the ball hits the ground \( 4 \) seconds after it was thrown.

a) Let:

\( d \): number of dimes and \( q \): number of quarters

Total number of coins is \(30\) \[ d + q = 30 \quad \text{(1)} \] A dime is worth \( 10 \) cents and a quarter is worth \( 25 \) cents, hence \[ 0.10d + 0.25q = 5.10 \quad \text{(2)} \] b) From equation (1): \( d = 30 - q \)

Substitute into equation (2): \[ 0.10(30 - q) + 0.25q = 5.10 \] Expand and simplify \[ 3 - 0.10q + 0.25q = 5.10 \] \[ 0.15q = 2.10 \] \[ q = 14 \] Then \( d = 16 \)

You have \( 16 \) dimes and \( 14 \) quarters.

a) The radius is distance from the center to any point on the circle, hence the use of the distance formula to find the radius \( r \) using the center \( (3,-2) \) and the point \( (6,2) \): \[ r = \sqrt{(6 - 3)^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \] b) The Equation of the circle is given by: \[ (x - 3)^2 + (y + 2)^2 = 25 \]

Factor all expressions** \[ x^2 - 9 = (x - 3)(x + 3)\] \[ x^2 - x - 6 = (x - 3)(x + 2) \] Now substitute in the given expression: \[ \frac{(x - 3)(x + 3)}{(x - 3)(x + 2)} \cdot \frac{x - 2}{x + 3} \] Cancel common factors \( x - 3 \) and \( x + 3 \) to obtain: \[ \frac{x^2 - 9}{x^2 - x - 6} \cdot \frac{x - 2}{x + 3} = \frac{x - 2}{x + 2} \] When we cancel common factors, as was done above, we are dividing by common factors. We cannot divide by zero, therefore we need to impose the conditions that the common factors canceled cannot be zero: \[ x - 3 \ne 0 \Rightarrow x \ne 3 \] \[ x + 3 \ne 0 \Rightarrow x \ne - 3 \] The given expression simplies to: \[ \frac{x - 2}{x + 2}, \quad \text{where } x \ne -3,\ -3 \]

Square both sides: \[ (\sqrt{2x + 3})^2 = (x - 1)^2 \Rightarrow 2x + 3 = x^2 - 2x + 1 \] Rearrange: \[ 0 = x^2 - 4x - 2 \] Use quadratic formula: \[ x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-2)}}{2(1)} = \frac{4 \pm \sqrt{16 + 8}}{2} \] \[ = \frac{4 \pm \sqrt{24}}{2} = \frac{4 \pm 2\sqrt{6}}{2} = 2 \pm \sqrt{6} \] Check for extraneous solutions:

Try \( x = 2 + \sqrt{6} \approx 4.45 \):

LHS (left hand side) = \( \sqrt{2(4.45) + 3} = \sqrt{11.9} \approx 3.45 \)

RHS (right hand side) = \( 4.45 - 1 = 3.45 \) ?

Try \( x = 2 - \sqrt{6} \approx -0.45 \):

LHS (left hand side) = \( \sqrt{2(-0.45) + 3} = \sqrt{-0.9 + 3} = \sqrt{2.1} \approx 1.45 \)

RHS (right hand side) = \( -0.45 - 1 = -1.45 \)

The solution \( x = 2 + \sqrt{6} \) verify both left and right sides of the given equation but \( x = 2 - \sqrt{6} \) does not and is therefore an extraneous solution.

The solution to the given equation is: \[ x = 2 + \sqrt{6} \]

Find slope of \( AB \) and \( CD \): \[ m_{AB} = \frac{7 - 3}{6 - 2} = \frac{4}{4} = 1 \] \[ m_{CD} = \frac{5 - 1}{10 - 6} = \frac{4}{4} = 1 \] Find slope of \( BC \) and \( AD \): \[ m_{BC} = \frac{5 - 7}{10 - 6} = \frac{-2}{4} = -0.5 \] \[ m_{AD} = \frac{1 - 3}{6 - 2} = \frac{-2}{4} = -0.5 \] Since both pairs of opposite sides are parallel, ABCD is a parallelogram.

First, express 625 as a power of 5: \[ 625 = 5^4 \] Use exponents rule to write \[ \dfrac{5^{2x}}{5} = 5^{2x-1} \] Rewrite the equation \[ 5^{2x - 1} = 5^4 \] which gives the algebraic equation \[ 2x - 1 = 4 \] Solve for \( x \) \[ x = \frac{5}{2} \]

Use Cosine Law: \[ c^2 = a^2 + b^2 - 2ab\cos(C) \] \[ c^2 = 7^2 + 10^2 - 2(7)(10)\cos(120^\circ) \] \[ c^2 = 49 + 100 - 140(-0.5) \] \[ c^2 = 149 + 70 = 219 \] \[ c = \sqrt{219} \approx 14.8 \ \text{cm} \] Hence: \[ c \approx 14.8 \ \text{cm} \]

Substitute \( x = 4 \), \( f(x) = 15 \) into the equation: \[ 15 = a(4 - 2)^2 + 3 \] Solve for \( a \) \[ 15 = a(2)^2 + 3 \] \[ a = 3 \] Function \( f(x) \) is a quadratic function given by: \[ f(x) = 3(x - 2)^2 + 3 \] From the form \( a(x - h)^2 + k \), the vertex is \[ (h, k) = (2, 3) \]

a) Let \( h \) be the height of the tower. Use the tangent formula in a right triangle: \[ \tan(32^\circ) = \frac{\text{h}}{50} \] Solve for \( h \): \[ h = 50 \cdot \tan(32^\circ) \approx 50 \cdot 0.6249 = 31.2\ \text{m} \] b) Height of the drone: \[ 31.2 + 30 = 61.2 \] Let \( \theta) \) be the angle of elevation to the drone: \[ \tan(\theta) = \frac{61.2}{50} \] Hence \[ \theta = \tan^{-1}(1.224) \approx 50.2^\circ \]

Let:

\( S \) be the speed of the boat in still water (in km/h),

\( r \) be the speed of the river current (in km/h),

\( d \) be the distance between points A and B (in km).

From the given: Downstream: \[ d = 3(S + r) \] Upstream: \[ d = 5(S - r) \] Equating both expressions for \( d \): \[ 3(S + r) = 5(S - r) \] Expand both sides: \[ 3S + 3r = 5S - 5r \] Put terms with \( r \) one side and terms with \( s \) on the other side: \[ 8 r = 2 s \] Solve for \( r \) \[ r = \frac{S}{4} \] Now substitute \( r = \frac{S}{4} \) back into the equation for downstream: \[ d = 3\left(S + \frac{S}{4}\right) = 3\left(\frac{5S}{4}\right) = \frac{15S}{4} \] Now, in still water (no current), the speed is just \( S \). Time to travel from A to B in still water: \[ \text{Time} = \frac{d}{S} = \frac{\frac{15S}{4}}{S} = \frac{15}{4} = 3.75 \text{ hours} \] So, the time it would take the boat to go from A to B in still water is: \[ 3 \text{ hours and } 45 \text{ minutes} \]

Expand the function: \[ f(x) = -3(x^2 - 14x + 40) = -3x^2 + 42x - 120 \] This is a quadratic function of the form: \[ f(x) = ax^2 + bx + c \quad \text{where } a = -3, \, b = 42 \] The function reaches its maximum at the vertex whose x coordinate is given by: \[ x = -\frac{b}{2a} = -\frac{42}{2(-3)} = \frac{42}{6} = 7 \] To find the maximum value, substitute \( x = 7 \) back into the function: \[ f(7) = -3(7 - 10)(7 - 4) = -3(-3)(3) = -3 \cdot -9 = 27 \] The function has a maximum at \( x = 7 \) and the maximum value of \( f(x) \) is \( f(7) = 27 \).

Let \( n \) be the number of students who scored below 60. and \( N \) be the number of students who scored 60 or more.

The average for students who scored below 60 is: \[ \frac{\sum X_i}{n} = 50 \Rightarrow \sum X_i = 50 n \] The average for students who scored 60 or more is: \[ \frac{\sum Y_i}{N} = 75 \Rightarrow \sum Y_i = 75 N \] The overall class average is 70: \[ \frac{\sum X_i + \sum Y_i}{20} = 70 \] Substitute: \[ \frac{50n + 75N}{20} = 70 \] Multiply all terms by \( 20 \) and simplify: \[ 50n + 75N = 1400 \quad (1) \] We also have: \[ n + N = 20 \Rightarrow N = 20 - n \] Substitute \( N \) into the equation (1): \[ 50n + 75(20 - n) = 1400 \] Expand and simplify: \[ -25n + 1500 = 1400 \] Solve for \( n \): \[ n = 4 \] \( n = 4 \) students who scored below 60.

Let \( h \) be the height of the trapezoid. \[ \text{Area} = \frac{1}{2} \cdot h (base_1 \; + \; base_2) = \frac{1}{2} \cdot h \cdot (10 + 10 + 3 + 4) = 270 \] Solve for \( h \): \[ h = 20 \] Apply the Pythagorean Theorem to the right triangle on the left (hypotenuse \( L \) ): \[ 20^2 + 3^2 = L^2 \quad \Rightarrow \quad L = \sqrt{409} \] Apply the Pythagorean Theorem to the right triangle on the right (hypotenuse \( R \) ): \[ 20^2 + 4^2 = R^2 \quad \Rightarrow \quad R = \sqrt{416} \] The perimeter of the trapezoid is: \[ \text{Perimeter} = \sqrt{409} + 10 + \sqrt{416} + 17 = 27 + \sqrt{409} + \sqrt{416} \; \text{units} \]

Let the length be \( L \) meters and the width be \( W \) meters such that \( L \gt W \). Use the area to write: \[ L \times W = 300 \] Use the perimeter to write: \[ 2L + 2W = 70 \] Divide all terms of the equation \( \; 2L + 2W = 70 \; \) by 2: \[ L + W = 35 \] Solve for \( L \): \[ L = 35 - W \] Substitute into the area equation: \[ (35 - W) \cdot W = 300 \] Expand: \[ 35W - W^2 = 300 \] Rearrange into a standard quadratic form: \[ W^2 - 35W + 300 = 0 \] Solve using the quadratic formula: \[ W = \frac{35 \pm \sqrt{(-35)^2 - 4(1)(300)}}{2(1)} \] \[ = \frac{35 \pm \sqrt{1225 - 1200}}{2} = \frac{35 \pm \sqrt{25}}{2} \] \[ W = \frac{35 \pm 5}{2} \Rightarrow W = 20 \text{ or } W = 15 \] For \( W = 15 \), \( L = 35 - 15 = 20 \).

For \( W = 20 \), \( L = 35 - 20 = 15 \).

Sine \( L \gt W \), the rectangle has dimensions \( L = 20 \) and \( W = 15 \).

The has a speed of \( 3000 \) revolutions per minute.

1 revolution is \( 360^\circ \)

1 minute is 60 second

Use abbreviations: min for minutes and rev for revolution to write: \[ \dfrac{3000 \text{rev}}{1 \text{min} } = \frac{3000 \, \text{rev}}{1 \, \text{min}} \times \frac{360^\circ}{1 \, \text{rev}} \times \frac{1 \, \text{min}}{60 \, \text{sec}} \] Cancel the units rev and min \[ = \frac{3000 \times 360}{60} \\ = \frac{1,080,000}{60} \\ = 18,000^\circ \, \text{per second} \] The motor rotates \( 18,000^\circ \) in one second.

Let \( x \) be the selling price of the house. \[ 6\% \text{ of } x = 8,880 \] \[ 0.06x = 8,880 \] Solve for \( x \) \[ x = \frac{8,880}{0.06} \] \[ x = 148,000 \] The selling price of the house was \( \$148,000 \).

We are given:

Rotational speed: \( 400 \, \text{rev/min} \)

Linear speed: \( 72 \, \text{km/h} = 72,000 \, \text{m/h} \)

First, convert revolutions per minute to revolutions per hour: \[ 400 \, \dfrac{\text{rev}}{\text{min}} \times 60 \, \dfrac{\text{min}}{\text{h}} = 24{,}000 \, \dfrac{\text{rev}}{\text{h}} \] Let \( C \) be the circumference of the tire in meters. The total distance traveled in one hour is also equal to the number of revolutions per hour multiplied by the circumference: \[ 24,000 \cdot C = 72,000 \] Solving for \( C \): \[ C = \frac{72,000}{24,000} = 3 \, \text{meters} \]

Let: \( x \) be the cost of one shirt , \( y \) be the cost of one pair of trousers and \( z \) the cost of one hat We are given: \[ \begin{aligned} \text{(1)} &\quad 4x + 4y + 2z = 560 \\ \text{(2)} &\quad 9x + 9y + 6z = 1290 \end{aligned} \] Divide equation (2) by 3: \[ \text{(3)} \quad 3x + 3y + 2z = 430 \] Now subtract equation (1) from equation (3): \[ (3x + 3y + 2z) - (4x + 4y + 2z) = 430 - 560 \] \[ - x - y = -130 \quad \Rightarrow \quad x + y = 130 \quad \text{(4)} \] Substitute equation (4) into equation (3): \[ 3(x + y) + 2z = 430 \] \[ 3(130) + 2z = 430 \] \[ 390 + 2z = 430 \] \[ 2z = 40 \quad \Rightarrow \quad z = 20 \] Now substitute back to find total cost of 1 shirt, 1 pair of trousers, and 1 hat: \[ x + y + z = 130 + 20 = 150 \] The total cost of one shirt, one pair of trousers, and one hat is \( \$150 \).

Let \( x \) represent the total number of toys.

The first child has \( \frac{x}{10} \) toys.

The second child has \( \frac{x}{10} + 12 \) toys.

The third child has \( \frac{x}{10} + 1 \) toys.

The fourth child has \( 2 \left( \frac{x}{10} + 1 \right) \) toys.

We set up the equation for the total number of toys: \[ \frac{x}{10} + \left( \frac{x}{10} + 12 \right) + \left( \frac{x}{10} + 1 \right) + 2 \left( \frac{x}{10} + 1 \right) = x \] Expand: \[ \frac{x}{10} + \frac{x}{10} + 12 + \frac{x}{10} + 1 + \frac{2 x}{10} + 2 = x \] Combine like terms: \[ \frac{6x}{10} + 15 = x \] Multiply through by 10 to eliminate the denominator: \[ 5x + 150 = 10x \] Subtract \( 5x \) from both sides: \[ 150 = 5 x \] Solve for \( x \): \[ x = \frac{150}{5} = 30 \] Thus, the total number of toys is \( 30 \).

First, we write each term in the product as a fraction: \[ \left( 1 - \frac{1}{10} \right) \left( 1 - \frac{1}{11} \right) \left( 1 - \frac{1}{12} \right) \cdots \left( 1 - \frac{1}{100} \right) \] This becomes: \[ \left( \frac{9}{10} \right) \left( \frac{10}{11} \right) \left( \frac{11}{12} \right) \cdots \left( \frac{99}{100} \right) \] Now, observe the cancellation pattern. Most terms cancel out, leaving: \[ \frac{9}{100} \] Thus, the final answer is: \[ \frac{9}{100} \]